dynamic regex in R

Question

The below code works so long as before and after strings have no characters that are special to a regex:

before <- 'Name of your Manager (note "self" if you are the Manager)' #parentheses cause problem in regex
after  <- 'CURRENT FOCUS'

pattern <- paste0(c('(?<=', before, ').*?(?=', after, ')'), collapse='')
ex <- regmatches(x, gregexpr(pattern, x, perl=TRUE))

Does R have a function to escape strings to be used in regexes?

Answer 1

In Perl, there is http://perldoc.perl.org/functions/quotemeta.html for doing exactly that. If the doc is correct when it says

Returns the value of EXPR with all the ASCII non-"word" characters backslashed. (That is, all ASCII characters not matching /[A-Za-z_0-9]/ will be preceded by a backslash in the returned string, regardless of any locale settings.)

then you can achieve the same by doing:

quotemeta <- function(x) gsub("([^A-Za-z_0-9])", "\\\\\\1", x)

And your pattern should be:

pattern <- paste0(c('(?<=', quotemeta(before), ').*?(?=', quotemeta(after), ')'),
                  collapse='')

Quick sanity check:

a <- "he'l(lo)"
grepl(a, a)
# [1] FALSE
grepl(quotemeta(a), a)
# [1] TRUE

Answer 2

Use \Q...\E to surround the verbatim subpatterns:

# test data
before <- "A."
after <- ".Z"
x <- c("A.xyz.Z", "ABxyzYZ")

pattern <- sprintf('(?<=\\Q%s\\E).*?(?=\\Q%s\\E)', before, after)

which gives:

> gregexpr(pattern, x, perl = TRUE) > 0
[1]  TRUE FALSE